A positive answer to the following question would be very helpful in understanding the evaluation of differential cohomology classes on chains.

Let $M$ be a smooth manifold and $c$ be a smooth $p$-chain in $M$, i.e. a finite linear combination of smoothly parametrized singular $p$ simplices. Let $|c|\subseteq M$ denote the support of the chain defined as a union of the simplices.

Here is the question: Does there exists an open subset $U\subseteq M$ such that
$|c|\subseteq U$ and $H^{q}(U,\mathbb{Z})=0$ for all $q\ge p+1$ (real coefficients would be sufficient in the application.

Here's an approach which might work (I'm not sure about the correctness of this.)

1) Assume $M$ is closed. Choose a triangulation $T$ of $M$.
If the support of $c$ is contained inside the $p$-skeleton of this triangulation, then
we take $U$ to be a small regular neighborhood of $p$-skeleton. This will do the job in this case.

2) If not, let's consider the dual triangulation $T^\ast$. We can ask whether or not
$c$ meets the $(n-p-1)$-skeleton of $T^\ast$. If it doesn't we can let $U$ be the effect
of deleting the $(n-p-1)$-skeleton of $T^\ast$ from $M$. Then it seems to me that
$U$ has the correct property in this case.

3 ) More generally, we can ask whether there exists a triangulation $T$
satisfying the (2). It seems to me that it should be possible to slightly modify $T$, say by general position, so that (2) holds.

Remark: I originally conceived of a version of this using Morse theory, but then realized that I had to retract it because I got confused. Perhaps it's possible to
find a Morse function $f\: M \to \Bbb R$ such that $c$ misses the $(n-p-1)$-skeleton
of the handlebody defined by $-f$? If so, we can define $U$ to be $M$ with this skeleton removed.

As the question was asked the answer seems to be: no. Consider R^n and a sequence of non-zero points converging to zero. Choose around each point in this sequence a small ball which does not meet the other points in the sequence and remove these closed balls. This is our manifold $M$. Now consider the 0-cycle given by the point 0. All neighbourhoods of 0 contain infinitely many holes and so $H^{n-1}(M)$ is non-zero.

If you ask the question, whether a given homology class has a representative with this property I agree with John that the answer should be yes. But I have not thought about a detailed argument.

Have you looked at the deformation theorem for rectifiable currents? This essentially states that any integral current $S$ can be approximated by a polyhedral current situated not very far from $S$. Your smooth chain defines an integral current. A good place to look for more details is Frank Morgan's Geometric Measure theory. A Beginner's Guide, Section 5.1. I believe that the strategy used in the proof of the deformation theorem could be useful for your problem too, or at least the weaker version suggested by Matthias Kreck.

More precisely, the deformation theorem indicates that your chain $c$ can be approximated (in various norms on the space of currents) by a nice polyhedral chain $c'$, whose support can be chosen in an arbitrarily small neighborhood of the support of $c$. In particular $c'$ is homologous to $c$, if $c$ is closed.

I tried similar ideas. Of course I could approximate my $c$ by a very regular chain or current $c^{\prime}$ which then has a neighbourhood as required. But this might be very small and may not contain the original $c$.
–
ubunkeJan 18 '13 at 22:14

The homology analogue of this question is answered affirmatively by Proposition 2.1 in math/0605535. The proof provides $U$ with $H_p(U;{\mathbb Z})$ torsion-free and so implies the affirmative answer to the original question by the Universal Coefficient Theorem.

The proof of Proposition 2.1 is a version of the argument John Klein described, phrased in terms of transverse triangulations of Lemma 2.3 (which is proved in math/1012.3979, after Matthias Kreck had pointed out that this had not been in the literature).